poj 2749 Building roads（2-sat）

Farmer John's farm has N barns, and there are some cows that live in each barn. The cows like to drop around, so John wants to build some roads to connect these barns. If he builds roads for every pair of different barns, then he must build N * (N - 1) / 2 roads, which is so costly that cheapskate John will never do that, though that's the best choice for the cows.

Clever John just had another good idea. He first builds two transferring point S1 and S2, and then builds a road connecting S1 and S2 and N roads connecting each barn with S1 or S2, namely every barn will connect with S1 or S2, but not both. So that every pair of barns will be connected by the roads. To make the cows don't spend too much time while dropping around, John wants to minimize the maximum of distances between every pair of barns.

That's not the whole story because there is another troublesome problem. The cows of some barns hate each other, and John can't connect their barns to the same transferring point. The cows of some barns are friends with each other, and John must connect their barns to the same transferring point. What a headache! Now John turns to you for help. Your task is to find a feasible optimal road-building scheme to make the maximum of distances between every pair of barns as short as possible, which means that you must decide which transferring point each barn should connect to.

We have known the coordinates of S1, S2 and the N barns, the pairs of barns in which the cows hate each other, and the pairs of barns in which the cows are friends with each other.

Note that John always builds roads vertically and horizontally, so the length of road between two places is their Manhattan distance. For example, saying two points with coordinates (x1, y1) and (x2, y2), the Manhattan distance between them is |x1 - x2| + |y1 - y2|.

The first line of input consists of 3 integers N, A and B (2 <= N <= 500, 0 <= A <= 1000, 0 <= B <= 1000), which are the number of barns, the number of pairs of barns in which the cows hate each other and the number of pairs of barns in which the cows are friends with each other.

Next line contains 4 integer sx1, sy1, sx2, sy2, which are the coordinates of two different transferring point S1 and S2 respectively.

Each of the following N line contains two integer x and y. They are coordinates of the barns from the first barn to the last one.

Each of the following A lines contains two different integers i and j(1 <= i < j <= N), which represent the i-th and j-th barns in which the cows hate each other.

The same pair of barns never appears more than once.

Each of the following B lines contains two different integers i and j(1 <= i < j <= N), which represent the i-th and j-th barns in which the cows are friends with each other. The same pair of barns never appears more than once.

You should note that all the coordinates are in the range [-1000000, 1000000].

You just need output a line containing a single integer, which represents the maximum of the distances between every pair of barns, if John selects the optimal road-building scheme. Note if there is no feasible solution, just output -1.

4 1 1
12750 28546 15361 32055
6706 3887
10754 8166
12668 19380
15788 16059
3 4
2 3

53246

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 1<<16
#define M 1<<16
#define inf 1<<30
int n,A,B;

////////////////////////////////////////////////////////
int tot;
int head
;
int vis
;
int tt;
int scc;
stack<int>s;
int dfn
,low
;
int col
;
struct Node
{
int from;
int to;
int next;
}edge[N<<3];
void init()
{
tot=0;
scc=0;
tt=0;
memset(head,-1,sizeof(head));
memset(dfn,-1,sizeof(dfn));
memset(low,0,sizeof(low));
memset(vis,0,sizeof(vis));
memset(col,0,sizeof(col));
while(!s.empty()){
s.pop();
}
}
void add(int s,int u)//邻接矩阵函数
{
edge[tot].from=s;
edge[tot].to=u;
edge[tot].next=head[s];
head[s]=tot++;
}
void tarjan(int u)//tarjan算法找出图中的所有强连通分支
{
dfn[u] = low[u]= ++tt;
vis[u]=1;
s.push(u);
int cnt=0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(dfn[v]==-1)
{
//    sum++;
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(vis[v]==1)
low[u]=min(low[u],dfn[v]);
}
if(dfn[u]==low[u])
{
int x;
scc++;
do{
x=s.top();
s.pop();
col[x]=scc;
vis[x]=0;
}while(x!=u);
}
}
bool two_sat(){

for(int i=0;i<2*n;i++){
if(dfn[i]==-1){
tarjan(i);
}
}
for(int i=0;i<n;i++){
if(col[2*i]==col[2*i+1]){
return false;
}
}
return true;
}
////////////////////////////////////////

int x
,y
;
int hate1
,hate2
;
int like1
,like2
;
int d
;
int D;
/////////////////////////////////////////////////
bool solve(int mid){
init();
for(int i=0;i<A;i++){//相互讨厌的建图
add(hate1[i]*2,hate2[i]*2+1);
add(hate2[i]*2,hate1[i]*2+1);
add(hate1[i]*2+1,hate2[i]*2);
add(hate2[i]*2+1,hate1[i]*2);
}
for(int i=0;i<B;i++){//相互喜欢的建图
add(like1[i]*2,like2[i]*2);
add(like2[i]*2,like1[i]*2);
add(like1[i]*2+1,like2[i]*2+1);
add(like2[i]*2+1,like1[i]*2+1);
}
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(d[i]+d[j]>mid){
add(i*2,j*2+1);
add(j*2,i*2+1);
}
if(d[i+n]+d[j+n]>mid){
add(i*2+1,j*2);
add(j*2+1,i*2);
}
if(d[i]+d[j+n]+D>mid){
add(i*2,j*2);
add(j*2+1,i*2+1);

}
if(d[i+n]+d[j]+D>mid){
add(j*2,i*2);
add(i*2+1,j*2+1);
}
}
}
if(two_sat()) return true;
return false;
}
////////////////////////////////////////////////
int main()
{
while(scanf("%d%d%d",&n,&A,&B)==3){

int sx1,sy1,sx2,sy2;
scanf("%d%d%d%d",&sx1,&sy1,&sx2,&sy2);

for(int i=0;i<n;i++){
scanf("%d%d",&x[i],&y[i]);
}
int num1,num2;
for(int i=0;i<A;i++){
scanf("%d%d",&hate1[i],&hate2[i]);
hate1[i]--;//记得--啊
hate2[i]--;

}
for(int i=0;i<B;i++){
scanf("%d%d",&like1[i],&like2[i]);
like1[i]--;//记得--啊
like2[i]--;
}
for(int i=0;i<n;i++){
d[i]=abs(x[i]-sx1)+abs(y[i]-sy1);
d[i+n]=abs(x[i]-sx2)+abs(y[i]-sy2);
}
D=abs(sx1-sx2)+abs(sy1-sy2);
int low=0;
int high=inf;
int ans=-1;//没有合适的方案输出-1，太粗心了
while(low<high){
int mid=(low+high)>>1;
//printf("%d\n",mid);
if(solve(mid)){
ans=mid;
high=mid;
}
else{

low=mid+1;
}
}
printf("%d\n",ans);

/* int l=0, r=8000000, mid, ans=-1;
while(l <= r){
mid = (l+r)>>1;
//buildGraph(mid);
if(solve(mid)){
ans = mid;
r = mid-1;
} else
l = mid+1;
}
printf("%d\n", ans);
*/
}
return 0;
}